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Karo-lina-s [1.5K]
3 years ago
11

Blanking on how medians in a triangle work

Mathematics
1 answer:
Sauron [17]3 years ago
4 0

A median cuts a leg of a triangle in half. So for instance, PF being a median means that DP and PE are congruent and have the same length. Consequently, each component triangle (that is, ∆DPC, ∆EPC, ∆EQC, etc) all have the same area.

If <em>A</em> is the area of ∆PCE, then the area of ∆PFE is 3<em>A</em>.

Take PF = 24 to be the base of ∆PFE, and take PC to be the base of ∆PCE. The area of any triangle is equal to 1/2 times the base times the height. But ∆PFE and ∆PCE share the same height (you can see this by dropping an altitude from the vertex E down to the line containing PF), so PC is 1/3 the length of PF, so FC must be 2/3 the length of PF.

So FC has length 2/3 × 24 = 16.

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8 0
3 years ago
All questions to be solved using linear combination.
monitta
1)
I:x-y=-7
II:x+y=7

add both equations together to eliminate y:
x-y+(x+y)=-7+7
2x=0
x=0

insert x=0 into II:
0+y=7
y=7

the solution is (0,7)

2)
I: 3x+y=4
II: 2x+y=5

add I+(-1*II) together to eliminate y:
3x+y+(-2x-y)=4+(-5)
x=-1

insert x=-1 into I:
3*-1+y=4
y=7

the solution is (-1,7)

3)

I: 2e-3f=-9
II: e+3f=18

add both equations together to eliminate f:
2e-3f+(e+3f)=-9+18
3e=9
e=3

insert e=3 into I:
2*3-3f=-9
-3f=-9-6
-3f=-15
3f=15
f=5

the solution is (3,5)

4)
I: 3d-e=7
II: d+e=5

add both equations together to eliminate e:
3d-e+(d+e)=7+5
4d=12
d=3

insert d=3 into II:
3+e=5
e=2

the solution is (3,2)

5)
I: 8x+y=14
II: 3x+y=4

add I+(-1*II) together to eliminate y
8x+y+(-3x-y)=14-4
5x=10
x=2

insert x=2 into II:
3*2+y=4
y=4-6
y=-2

the solution is (2,-2)
8 0
4 years ago
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